Sidebilder
PDF
ePub

Schol. It is plain (APP. I. 7, schol.) that E is the point in which the straight lines drawn from the angles of the triangle ABC bisecting the opposite sides would cut one another.

It may be farther remarked, that this porism may be regarded as the case in which the following problem becomes indeterminale :

Through a given point to draw a straight line, such that the perpendicular to it from one of three given points may be equal to the sum of those drawn to it from the other two;"- -a problem which is always determinate, except when the point through which the line is to be drawn, is the same as E, deterinined as above; since in every other case E is to be found in the manner already shown, and the required line is drawn through it and the given point.

PROP. XVII. PORISM.—A circle and a straight line being given, a point may be found within the circle, such that if any chord be drawn through it, the centre of the circle which passes through the extremities of the chord, and through the centre of the given circle, will be in the given straight line.

Let ABC be a given circle, and DE a straight line given in position; a point F may be found within the circle, such that the centre of a circle described through G, the given centre, and through the extremities of any chord AC, passing through F, will be in DE.

Draw GE perpendicular to DE, and take FG such that 2EG : BG :: BG : GF; F is the point through which the chords pass.

Through F let any chord AC, be drawn, and draw GHD perpendicular to it, and join AG. Now (III. 31, and VI. 8, cor.) the square of AG is equal to the rectangle under GH, and twice the radius of a circle passing through A, G, C. But (const.) BG? or AGis equal to 2EG.GF; and (VI. 4) DG:GE:: FG : GH; whence (VI. 16, and by doubling) 2DG.GH is equal to 2EG.GF, and (I. ax. 1) 2DG.GH is equal to AGʻ. Hence the rectangle under GH and twice DG is equal to the rectangle under GH and twice the radius of the circle passing through A, G, C: therefore DG must be the radius of that circle, and its centre is therefore in DE.

Schol. The property developed in this proposition, might also be expressed in form of the following local theorem: The locus of the centre of a circle described through the centre of a given circle, and the extremities of a chord passing through a given point, is a straight line given in position.

D

B

E

B

F

G

Prop. XVIII. THEOR.-Of isoperimetrical triangles on the same base, the isosceles one is the greatest.

K

D

E

Let ABC be an isosceles triangle on the base BC ; it is greater than any other triangle on BC, of equal perimeter, and not isosceles; that is, than any other triangle on BC, having the sum of its other sides equal to the sum of AB, AC.

Draw DAE parallel and AF perpendicular to BC: the altitude of any triangle on BC, having its perimeter equal to that of ABC, and not being isosceles, is less than AF. For, through any other point G, in the parallel draw GB, GC. Then the angles which BA, CÁ make with DE are equal, being (I. 29) equal to the equal angles ABC, ACB; while those which GB, GC make with it are unequal, being equal to the unequal angles "GBC, GCB. Hence (APP. I. 12) the sum of BG, GC is greater than the sum of BA, AC. Let GH be perpendicular to BC, and draw BK, CK to any point K in its continuation through G. Then (I. 19) in the triangles BGK, CGK, BK is greater than BG, and CK than CG; wherefore BK, KC are greater than BG, GC, and consequently than BA, AC, which are less than BG, GC. Hence, by adding BC, the triangle BGC, which has its altitude equal to AF; and the triangle BKC, which has its altitude greater than AF, have each a greater perimeter than ABC: and therefore a triangle on BC, which is not isosceles, and which has its perimeter equal to that of ABC, must have its altitude less than AF, and must therefore be less than ABC.*

B

F

[ocr errors]

PROP. XIX. THEOR.-If two sides of a triangle be given, the triangle is a maximum, when they contain a right angle.

Let ABC, ABD be triangles on the base AB, and having the sides BC, BD equal: if ABC be a right angle, the triangle ABC is greater than ABD. For (I. 19) the perpendicular DE is less than DB, or than the perpendicular BC ; and therefore, since the base AB is common, the triangle ABD is less than ABC.

[ocr errors]

A

Prop. XX. THEOR.-Of isoperimetrical polygons, having a given number of sides, the maximum is the one which is equilateral

* The demonstration of this proposition given in Legendre's Geometry, Book IV. Appendix, is deficient, as it is assumed without proof, that, in his diagram, the point M is not in the straight line AN.

F

E

[ocr errors]

For, if possible, let ABCDE be the maximum polygon, and yet the side AE greater than AB. On BE describe the isosceles triangle BFE having the sum of its sides BF, FE equal to the sum of BA, AE. Then (APP. II. 18) the triangle BFE is greater than BAE, and (I. ax. 4) the polygon FBCDE than ABCDE. Hence the latter is not a maximum, and the same may be shown in a similar manner, unless the sides be all equal.

Schol. Hence one requisite in constituting a polygon, of a given perimeter and a given number of sides, a maxiinum, is that it must be equilateral.

B

Prop. XXI. THEOR.-If all the sides of a quadrilateral or polygon except one be given in magnitude, the figure will be a maximum, when that remaining side is the diameter of a semicircle, and the figure is inscribed in that semicircle.

Let ABCDE be the greatest polygon that can be contained by sides AB, BC, CD, DE given in magnitude, and AE not given: AE is the diameter of a semicircle, and the angular points of the polygon are on the arc of that semicircle.

Soin BE. Then, if ABCDE be not a semicircle, and therefore ABE not a right angle, by making it a right angle, the triangle ABE (APP. II. 19) would be enlarged, as would also the whole polygon, the magnitude of the part BCDE not being changed by a change in the angle ABE. If the polygon be a maximum, therefore, ABE must be a right angle, and the segment ABE a semicircle. In the same manner it would be shown, by drawing lines from C and D to A and E, that ACE, ADE are right angles; and the proof will be the same if there be six or any greater number of sides; whence the truth of the proposition is manifest.

Schol. There is but one semicircle which will contain the maximum polygon. For, suppose ABCDE to be a semicircle containing it;

the angles at its centre subtended by the chords AB, BC, CD, DE, amount to two right angles, which would no longer be the case, were the radius made either greater or less ; as the triangles having their vertices at the centre, and contained by the radii and the sides AB, BC, &c. would evidently have the magnitudes of their angles changed.

Schol. It is plain that the area of the polygon will be the same in whatever order the sides are arranged; as, however they are placed, they will cut off equal segments, and the polygon is the part of the semicircle remaining, when those segments are taken away.

A

B

Prop. XXII. THEOR.-Of quadrilaterals and polygons which have their sides equal, each to each, the greatest is that which can be inscribed in a circle.

In the rectilineal figures ABCDE, FGHKL, let the sides AB, BC, CD, DE, EA be severally equal to FG, GH, HK, KL, LF, and let AD be inscribed in a circle, but FK not be capable of being inscribed in one: AD is greater than FK.

Draw the diameter CM, and join AM, EM; construct the triangle FNL, having the sides FN, NL respectively equal to AM, ME," so that (I. 8) the triangles themselves are equal, and join HN. Then, by the preceding proposition, the figures CBAM, CDEM are respectively greater than HGFN, HKLN; and therefore (I. ax. 4) the polygon ABCDEM is greater than FGHKLN; and, by taking away the equal triangles AME, FNL, there remains the figure ABCDÉ greater than FGHKL.

Cor. From this proposition and the twentieth it follows, that of polygons of the same number of sides, and of equal perimeters, the regular polygon is the greatest ; and that of quadrilaterals of equal perimeters, the greatest is the square.

M

H

N

E

D

h

Prop. XXIII. THEOR.-Of regular polygons which have equal perimeters, that which has the greater number of sides is the greater.

Let AB be half the side of the polygon which has the less number of sides, and BC a perpendicular to it, which will evidently pass through the centre of its inscribed or circumscribed circle set C be that centre, and join AC. Then, ACB will be the angle at the centre subtended by the half side AB.

Make BCD equal to the angle subtended at the centre of the other polygon by half its side, and from C as centre, with CD as radius, describe an arc cutting AC in E, and CB produced in F. Then, it is plain, that the angle ACB is to four right angles, as AB to the common perimeter; and four right angles are to DCB, as the common perimeter to the half of a side of the other polygon, which, for brevity, call S: then, ex æquo, the angle ACB is to DCB, as AB to S. But (VI. 33) the angle ACB is to DCB, as the sector ECF to the sector DCF; and consequently (V. 11) the sector ECF is to DCF, as AB to S, and, by_division, the sector ECD is to DCF, as AB-S tó s. Now the triangle ACD is greater than the sector CED, and DCB is less than DCF. But (VI. 1) these triangles are as their bases AD, DB; therefore AD has to DB a greater ratio, than AB-S to S. Hence AB, the

H

E

[ocr errors]

GD

B

sum of the first and second, has to DB, the second, a greater ratio than AB, the sum of the third and fourth, has to s the fourth ;* and therefore (V. 10) S is greater than DB. Let then BG be equal to S, and draw GH parallel to DC, meeting FC_produced in H, Then, since the angles GHB, DCB are equal, BH is the perpendicular drawn from the centre of the polygon having the greater number of sides to one of the sides; and since this is greater than BC, the like perpendicular in the other polygon, while the perimeters are equal, it follows that the area of that which has the greater number of sides is greater than that of the other,

PROP. XXIV. THEOR. A circle is greater than any regular polygon of the same perimeter.

For, if a polygon P similar to the one Q proposed for comparison, be described about the circle C: since the area of the circumscribed polygon is evidently equal to the rectangle under its perimeter and half the radius, and (APP. I. 39) the area of the circle equal to the rectangle under its perimeter and half its radius ; therefore (VI. 1) P is to C, as the perimeter of P to the perimeter of C, or of Q: and (VI. 20) P is to Q in the duplicate ratio of a side of P to a side of Q, or (VI. 20, cor. 4) in the duplicate ratio of the perimeter of P to the perimeter of Q. Hence P is to a mean proportional between P and Q, as the perimeter of P to the perimeter of Q; and therefore (V. 11) P is to C, as P to a mean proportional between P and Q; that is, (V. 9) C is a mean proportional between P and Q; and (V. 14) being less than P, it is greater than Q, the polygon of equal perimeter.

Schol. Every thing here proved will evidently hold equally regarding a círcle C, as compared with any polygon Q of equal perimeter, which is such that a circle can be inscribed in it; for then a polygon P, similar to Q, can be described about C.

Cor. Hence it appears also, that a circle is a mean proportional between any rectilineal figure described about it, and a similar one of equal perimeter with the circle.

* This may be proved by dividing the first term by the second, and the third by the fourth, and adding unity to each quotient. See the Supplement to Book V.

« ForrigeFortsett »